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3 years ago

What is the sign of enthalpy change for (a), an endothermic reaction? (b). an exothermic reaction?

Chemistry

1 answer:

fenix001 [56]3 years ago

8 0

Answer:

The answer is: (a) positive; (b) negative.

Explanation:

The change in enthalpy (ΔH) of a reaction is the amount of energy absorbed or released during a chemical reaction carried out at constant pressure.

a) In an endothermic chemical reaction, heat energy is absorbed by the system from the surrounding. Therefore, the sign of enthalpy change for an endothermic process is positive, ΔH= positive.

b) In an exothermic chemical reaction, heat energy is released by the system into the surrounding. Therefore, the sign of enthalpy change for an exothermic process is negative, ΔH= negative.

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First, we will calculate the partial pressure of SF₆ using the following expression.

$pSF_6 = P \times \frac{ppb}{10^{9} }$

where,

pSF₆: partial pressure of SF₆

P: total pressure of air (we will assume it is 1 atm)

ppb: concentration of SF₆ in parts per billion

$pSF_6 = P \times \frac{ppb}{10^{9} } = 1 atm \times \frac{1.0 ppb}{10^{9} } = 1.0 \times 10^{-9} atm$

Then, we will convert 1.0 cm³ to L using the following conversion factors:

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$1.0 cm^{3} \times \frac{1mL}{1cm^{3}} \times \frac{1L}{1000 mL} = 1.0 \times 10^{-3} L$

Next, we will convert 46 °C to Kelvin using the following expression.

$K = \° C + 273.15 = 46 + 273.15 = 319 K$

Afterward, we calculate the moles (n) of sulfur hexafluoride using the ideal gas equation.

$P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T} = \frac{1.0 \times 10^{-9} atm \times 1.0 \times 10^{-3} L}{(0.082 atm.L/mol.K) \times 319 K} = 3.8 \times 10^{-14} mol$

Finally, we will convert 3.8 × 10⁻¹⁴ mol to molecules using Avogadro's number.

$3.8 \times 10^{-14} mol \times \frac{6.02 \times 10^{23} molecules }{mol} = 2.3 \times 10^{10} molecules$

You can learn more about partial pressure here: brainly.com/question/13199169

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